Evaluating the series $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} $

Wolfram MathWorld states that $$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3) \, , $$

where $\psi_{1}(x)$ is the trigamma function.

I can’t seem to get the answer in that form.

Using the Taylor expansion $ \displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$, I get

$$ \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, dx. $$

Then I integrating by parts, I get

$$ \begin{align} &4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, dx \\ &= \frac{\pi^{2}}{9} \ln \left(\frac{1}{2} \right) – 8 \int_{0}^{\frac{1}{2}} \frac{\arcsin (x) \ln (x)}{\sqrt{1-x^{2}}} \, dx \\ &= – \frac{\pi^{2}}{9} \ln 2 – 8 \int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, du \\ &= – 8 \ln 2 \int_{0}^{\frac{\pi}{6}} u \, du – 8\int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, du \\ &= – 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \, du \\ &= -8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \ln (1-e^{2iu}) \, du \\ &= 8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \sum_{n=1}^{\infty} \frac{e^{2in u}}{n} \, du \\ &= 8 \ \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\frac{\pi}{6}} u \cos (2nu) \, du \\ &= \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} – 2 \zeta(3) \\ &= \frac{2 \pi}{3} \Bigg( \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{2}} + \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{2}} – \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{2}} – \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{2}} \Bigg) \\ &+ \ 2 \Bigg( \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{3}} – \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{3}} – \sum_{n=0}^{\infty} \frac{1}{(6n+3)^{3}} – \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{3}} \\ &+ \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{3}} + \sum_{n=1}^{\infty} \frac{1}{(6n)^{3}} \Bigg) – 2 \zeta(3) \\ &= \frac{\pi \sqrt{3}}{108} \Bigg( \psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) – \psi_{1}\left(\frac{5}{6} \right) \Bigg) + \frac{1}{432} \Bigg( – \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &-28 \zeta(3) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) + 4 \zeta(3)\Bigg) – 2 \zeta (3) .\end{align}$$

From here I’ve been going in circles trying to get the answer in the form given above.


EDIT:

Using the duplication formula for the trigamma function (i.e., $ \displaystyle 4 \psi_{1}(2x) = \psi_{1}(x) + \psi_{1} \left(x + \frac{1}{2} \right) $),

$$ \begin{align} &\psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) – \psi_{1}\left(\frac{5}{6} \right) \\ &= 4 \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) – 4 \psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) \\ &= 6 \psi_{1} \left(\frac{1}{3} \right) – 6 \psi_{1} \left(\frac{2}{3} \right). \end{align}$$

Therefore,

$$ \begin{align} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &= \frac{\sqrt{3} \pi}{18} \Bigg( \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Bigg) +
\frac{1}{432} \Bigg( – \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &+ \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) – 2 \zeta(3)\Bigg) – 2 \zeta (3). \end{align}$$

So it comes down to somehow showing that $$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3) .$$

SECOND EDIT:

Using the duplication formula for $\psi_{2}(x)$, we get

$$ \begin{align} &-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) \\ &= -8 \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) – 8 \psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &= -6 \psi_{2} \left(\frac{1}{3} \right) – 6 \psi_{2} \left(\frac{2}{3} \right) . \end{align}$$

And using the more general multiplication formula, we get

$$ \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) + \psi_{2}(1) = 27 \psi_{2} (1) .$$

Therefore,

$$ -6 \psi_{2} \left(\frac{1}{3} \right) – 6 \psi_{2} \left(\frac{2}{3} \right) = -156 \psi_{2} (1) = 312 \zeta(3) .$$

Solutions Collecting From Web of "Evaluating the series $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} $"

Using the Taylor expansion for $\arcsin^2(x)$ and integrating, it’s easy to show (as you made) that
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}}
&= 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \operatorname{d}x = – 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \operatorname{d}u \\
&= \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} – 2 \zeta(3)\\
&=\frac{2 \pi}{3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+2\operatorname{Cl}_3\left(\frac{\pi}{3}\right)-2 \zeta(3)
\end{align}
$$
recalling that the Clausen function are defined as
$$
\operatorname{Cl}_{m}(\theta)=
\begin{cases}\displaystyle
\sum_{n=1}^{\infty} \frac{\sin (n\theta)}{n^{m}}& \text{for } m\text{ even}\\
\displaystyle\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{n^{m}}& \text{for } m\text{ odd}
\end{cases}
$$
The value of the Clausen function $\operatorname{Cl}_3$ at $\frac{\pi}{3}$ is
$$
\operatorname{Cl}_3\left(\frac{\pi}{3}\right)=\frac{1}{2}\left(1-2^{-2}\right)\left(1-3^{-2}\right)\zeta(3)=\frac{1}{3}\zeta(3)
$$
From the duplication formula
$$
\operatorname{Cl}_{2m}(2\theta)=2^{2m-1}\left[\operatorname{Cl}_{2m}(\theta)-\operatorname{Cl}_{2m}(\pi-\theta)\right]
$$
we find
$$
\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)=\frac{2}{3}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
\begin{align}
\psi_1\left(\frac{1}{3}\right) &=\frac{2\pi^2}{3}+3\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)\\
\psi_1\left(\frac{2}{3}\right) &=\frac{2\pi^2}{3}-3\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)
\end{align}
$$
one has
$$
\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)=6\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)
$$
and then
$$
\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)=\frac{\sqrt 3}{12}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]
$$

Finally, putting all together, we have
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}}
&=\frac{2 \pi}{3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+2\operatorname{Cl}_3\left(\frac{\pi}{3}\right)-2 \zeta(3)\\
&=\frac{2 \pi}{3}\frac{\sqrt 3}{12}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]+2\frac{1}{3}\zeta(3)-2\zeta(3)\\
&=\frac{\pi\sqrt 3}{18}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]-\frac{4}{3}\zeta(3).
\end{align}
$$

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$\ds{\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3)}$

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n{2n \choose n}}
=\sum_{n = 1}^{\infty}x^{n}\,
{\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
=\sum_{n = 1}^{\infty}x^{n}\int_{0}^{1}t^{n – 1}\pars{1 – t}^{n}\,\dd t
\\[3mm]&=\int_{0}^{1}\sum_{n = 1}^{\infty}\bracks{tx\pars{1 – t}}^{n}
\,{\dd t \over t}
=\int_{0}^{1}\bracks{{1 \over 1 – tx\pars{1 – t}} – 1}\,{\dd t \over t}
=\int_{0}^{1}{x\pars{1 – t} \over 1 – tx\pars{1 – t}}\,\dd t
\end{align}

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}}
=\int_{0}^{x}\dd y\int_{0}^{1}{1 – t \over 1 – ty\pars{1 – t}}\,\dd t
=-\int_{0}^{1}{\ln\pars{1 – \bracks{1 – t}tx} \over t}\,\dd t
\end{align}

\begin{align}
&\color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over n^{3}{2n \choose n}}}
=-\int_{0}^{1}{\dd x \over x}
\int_{0}^{1}{\ln\pars{1 – \bracks{1 – t}tx} \over t}\,\dd t
=\int_{0}^{1}{{\rm Li}_{2}\pars{t\bracks{1 – t}} \over t}\,\dd t
\\[3mm]&=\color{#66f}{{\pi \over 36\root{3}}\, \left[\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right]-\frac{4 \zeta (3)}{3}}
\approx {\tt 0.5229}
\end{align}